Question

Find the number of factors of $192$ which are multiples of both $6$ and $8.$

Answer 1

Least common multiple of 6 and 8 is $2\times 2\times 2\times 3=24$

Hence, the number which is multiple of both $6$ and $8$ must be a multiple of $24$ also.

We need to find factors of $192$ which are multiple of $24$

To figure out factors of $192,$ divide it by $24.$
$\frac{192}{24}=8$
$\Rightarrow 24\times 8=192$

$192=24\times 2\times 2\times 2$
Hence, multiples of $24$ and factors of 192 are:
$24,24\times 2,24\times 4,24\times 8$

$\therefore$ There are $4$ multiples of both $6$ and $8,$ which are also a factor of $192.$